YES
The TRS could be proven terminating. The proof took 225 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (4ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (194ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| app#(app(plus, app(s, x)), y) | → | app#(s, app(app(plus, x), y)) | | app#(app(plus, app(s, x)), y) | → | app#(plus, x) |
| app#(app(plus, app(s, x)), y) | → | app#(app(plus, x), y) |
Rewrite Rules
| app(id, x) | → | x | | app(plus, 0) | → | id |
| app(app(plus, app(s, x)), y) | → | app(s, app(app(plus, x), y)) |
Original Signature
Termination of terms over the following signature is verified: id, app, plus, 0, s
Strategy
The following SCCs where found
| app#(app(plus, app(s, x)), y) → app#(app(plus, x), y) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| app#(app(plus, app(s, x)), y) | → | app#(app(plus, x), y) |
Rewrite Rules
| app(id, x) | → | x | | app(plus, 0) | → | id |
| app(app(plus, app(s, x)), y) | → | app(s, app(app(plus, x), y)) |
Original Signature
Termination of terms over the following signature is verified: id, app, plus, 0, s
Strategy
Polynomial Interpretation
- 0: 0
- app(x,y): y + x
- app#(x,y): y + 2x
- id: 0
- plus: 0
- s: 1
Improved Usable rules
| app(app(plus, app(s, x)), y) | → | app(s, app(app(plus, x), y)) | | app(plus, 0) | → | id |
| app(id, x) | → | x |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| app#(app(plus, app(s, x)), y) | → | app#(app(plus, x), y) |